There are times when an aircraft will have a weight and
balance calculation done, known as an extreme condition
check. This is a pencil and paper check in which
the aircraft is loaded in as nose heavy or tail heavy a
condition as possible to see if the center of gravity will
be out of limits in that situation. In a forward adverse
check, for example, all useful load in front of the forward
CG limit is loaded, and all useful load behind this
limit is left empty. An exception to leaving it empty
is the fuel tank. If the fuel tank is located behind the
forward CG limit, it cannot be left empty because the
aircraft cannot fly without fuel. In this case, an amount
of fuel is accounted for, which is known as minimum
fuel. Minimum fuel is typically that amount needed
for 30 minutes of flight at cruise power.

For a piston engine powered aircraft, minimum fuel
is calculated based on the METO (maximum except
take-off) horsepower of the engine. For each METO
horsepower of the engine, one-half pound of fuel is
used. This amount of fuel is based on the assumption
that the piston engine in cruise flight will burn 1 lb
of fuel per hour for each horsepower, or 1/2 lb for 30
minutes. The piston engines currently used in small
general aviation aircraft are actually more efficient than
that, but the standard for minimum fuel has remained
the same.

For example, if a forward adverse condition check was
being done on a piston engine powered twin, with each
engine having a METO horsepower of 500, the minimum
fuel would be 250 lb (500 METO Hp ÷ 2).

For turbine engine powered aircraft, minimum fuel is
not based on engine horsepower. If an adverse condition
check is being performed on a turbine engine
powered aircraft, the aircraft manufacturer would need
to supply information on minimum fuel.

Tare Weight

When aircraft are placed on scales and weighed, it
is sometimes necessary to use support equipment to
aid in the weighing process. For example, to weigh a
tail dragger airplane, it is necessary to raise the tail in
order to get the airplane level. To level the airplane,
a jack might be placed on the scale and used to raise the tail. Unfortunately, the scale is now absorbing the
weight of the jack in addition to the weight of the
airplane. This extra weight is known as tare weight,
and must be subtracted from the scale reading. Other
examples of tare weight are wheel chocks placed on
the scales and ground locks left in place on retractable
landing gear.

Procedures for Weighing an Aircraft

General Concepts

The most important reason for weighing an aircraft
is to find out its empty weight (basic empty weight),
and to find out where it balances in the empty weight
condition. When an aircraft is to be flown, the pilot in
command must know what the loaded weight of the
aircraft is, and where its loaded center of gravity is. In
order for the loaded weight and center of gravity to be
calculated, the pilot or dispatcher handling the flight
must first know the empty weight and empty weight
center of gravity.

Earlier in this chapter it was identified that the center
of gravity for an object is the point about which the
nose heavy and tail heavy moments are equal. One
method that could be used to find this point would
involve lifting an object off the ground twice, first
suspending it from a point near the front, and on the
second lift suspending it from a point near the back.
With each lift, a perpendicular line (90 degrees) would
be drawn from the suspension point to the ground. The
two perpendicular lines would intersect somewhere in
the object, and the point of intersection would be the
center of gravity. This concept is shown in Figure 4-4,
where an irregular shaped object is suspended from
two different points. The perpendicular line from the
first suspension point is shown in red, and the new suspension point line is shown in blue. Where the red
and blue lines intersect is the center of gravity.

If an airplane were suspended from two points, one
at the nose and one at the tail, the perpendicular drop
lines would intersect at the center of gravity the same
way they do for the object in Figure 4-4. Suspending
an airplane from the ceiling by two hooks, however,
is clearly not realistic. Even if it could be done, determining
where in the airplane the lines intersect would
not be possible.

A more realistic way to find the center of gravity for an
object, especially an airplane, is to place it on a minimum
of two scales and to calculate the moment value for each
scale reading. In Figure 4-5, there is a plank that is 200"
long, with the left end being the datum (zero arm), and
6 weights placed at various locations along the length of
the plank. The purpose of Figure 4-5 is to show how the
center of gravity can be calculated when the arms and
weights for an object are known.

To calculate the center of gravity for the object in
Figure 4-5, the moments for all the weights need to
be calculated and then summed, and the weights need
to be summed. In the four column table in Figure 4-6,
the item, weight, and arm are listed in the first three
columns, with the information coming from Figure
4-5. The moment value in the fourth column is the product of the weight and arm. The weight and moment
columns are summed, with the center of gravity being
equal to the total moment divided by the total weight.
The arm column is not summed. The number appearing
at the bottom of that column is the center of gravity.
The calculation would be as shown in Figure 4-6.

For the calculation shown in Figure 4-6, the total
moment is 52,900 in-lb, and the total weight is 495 lb.
The center of gravity is calculated as follows:

An interesting characteristic exists for the problem presented
in Figure 4-5, and the table showing the center
of gravity calculation. If the datum (zero arm) for the
object was in the middle of the 200" long plank, with
100" of negative arm to the left and 100" of positive
arm to the right, the solution would show the center
of gravity to be in the same location. The arm for the
center of gravity would not be the same number, but
its physical location would be the same. Figure 4-7 and
Figure 4-8 show the new calculation.

In Figure 4-7, the center of gravity is 6.9" to the right
of the plank’s center. Even though the arm is not the
same number, in Figure 4-5 the center of gravity is also
6.9" to the right of center (CG location of 106.9 with the
center being 100). Because both problems are the same
in these two figures, except for the datum location, the
center of gravity must be the same.

The definition for center of gravity states that it is the
point about which all the moments are equal. We can
prove that the center of gravity for the object in Figure
4-7 is correct by showing that the total moments on
either side of this point are equal. Using 6.87 as the
CG location for slightly greater accuracy, instead of
the rounded off 6.9 number, the moments to the left of
the CG would be as shown in Figure 4-9.

The moments to the right of the CG, as shown in Figure
4-7, would be as shown in Figure 4-10.

Disregarding the slightly different decimal value,
the moment in both of the previous calculations is
10,651 in-lb. Showing that the moments are equal is
a good way of proving that the center of gravity has
been properly calculated.